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:: 1 ::Sr. INTER IIT-JEE MATERIAL

CHEMISTRY STOICHIOMETRYv v v v v v v v v v v v v v v v v v v v v v v v

Synopsis...

PART - I
(Laws of chemical combination, amu, Atomic and molecular masses, average

atomic mass, gm. atom & gm.molecule)

Laws of Chemical combination :
1. Law of conservation of Mass (Lavoisier, 1744)

“Mass can not be created or destroyed. In physical or chemical process, the total mass of the system
remain conserved.”

 This law can not be applied to the nuclear process where mass and energy are interconversable.

(Einstein’s equation : 2.E m C  )

 On the basis of this law, we may conclude that for a reaction, if the reaction is 100% completed then, total
mass of reactants before reaction = Total mass of products after reaction.

 For incomplete reactions :
Total mass of reactants before reation = Total mass of products formed + mass of unreacted reactants left.

CAPS - 1 : .3.4g of AgNO
3
in 100g water, when mixed with 1.17g of NaCl in 100g water, 2.87g AgCl

and 1.70g NaNO
3
were obtained. Verify law of conservation of mass.

Sol : Total mass of substance before reaction
= 3.4g AgNO

3
+ 100g H

2
O + 1.17g NaCl + 100g H

2
O

= 204.57g of reactant

 Total mass of substance after reaction
2.87g AgCl + 1.70g NaNO

3
+ 200g H

2
O

= 204.57g of Products
 The result proves law of conservation of mass.
Law of constant (or definite) proportions (Proust, 1799)

“A chemical compound always contains the same element combined together in the same proportions by
mass.”

 i.e, the composition of a compound always remain fixed and it is independent to the source from which the
compound is obtained.

 Eg : Compound CO
2
can be formed by either of these process.

i) by heating CaCO
3

3 2CaCO CaO CO


 

ii) by heating NaHCO
3

3 2 3 2 22NaHCO Na CO H O CO


  

iii) by burning in O
2

2 2C O CO


 

iv) by reaction of CaCO
3
with HCl.


3 2 2 22CaCO HCl CaCl H O CO   

 CO
2
obtained by all these methods contains C:O ratio 12 : 32 by mass.

 This law can not be applied to the compound, obtained by using different isotopes of the elements as the
isotopes have different atomic masses.
Eg : CO

2
using C12 isotope has C : O :: 12 : 32

CO
2
using C14 isotope has C : O :: 14 : 32

 The elements combining in the same ratio of their masses may give different compounds under different

STOICHIOMETRY

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experimental conditions.
Eg : Combination of ‘C’, ‘H’ and ‘O’ in the ratio 12 : 3 : 8 may give C

2
H

5
OH or CH

3
OCH

3
under different

experimental conditions.
CAPS - 2 : .A sample of 1.375g cupric oxide when reduced in a steam of hydrogen gave 1.098g CU.

On the other hand, a sample of 1.179g pure CU gave 1.476g cupric oxide when CU was in HNO
3

and nitrate formed was heated strongly to get cupric oxide. Show that these data prove law of
definite proportions.

Sol : According to Ist experiment, the composition of cupric oxide is
1.375g cupric oxide = 1.098g CU + Oxygen (0.277g)

 % of CU in the sample =
1.098

100 79.85%
1.375

 

 In 2nd experiment, the composition of cupric oxide is :
1.476g cupric oxide = 1.179g CU + oxygen (0.297g)

 % of CU in the sample =
1.179

100 79.87%
1.476

 

 Since, in both the experiments % CU and % O are constant which validate law of constant proportions
Law of Multiple proportions (Dalton)

“If two elements combine to form more than one compound, then for the fixed mass of one element, the
mass of other element combined will be in simple ratio”

 Examples of law of multiple proportions :
i) Combination of C and O may form CO and CO

2

In CO ratio of C : O is 12 : 16
In CO

2
ratio of C : O is 12 : 32

Thus ratio ‘O’ in CO and CO
2
is 16 : 32 or 1 : 2 i.e, a whole number ratio.

ii) N and O form five stable oxides, N
2
O, NO, N

2
O

3
, N

2
O

4
and N

2
O

5
. In these oxides, amount of oxygen,

which react with 28g N
2
are in the ratio

16 : 32 : 48 : 64 : 80. i.e, 1 : 2 : 3 : 4 : 5.
iii) In H

2
O and H

2
O

2
, 2g H combines with 16g and 32 g oxygen respectively and the ratio is 16 : 32 or 1 : 2.

 However, the discovery of isotopes led to some disperencies in this law also.
CAPS-3 : Two compounds each containing only tin and oxygen had the following composition.

Mass Mass
% of tin % of oxygen

Compound A 78.77 21.23
Compound B 88.12 11.88
Verify law of multiple proportions for this data.

Sol : In compound A
21.23 parts of oxygen combine with 78.77 parts of tin
1 part of oxygen combines with 78.77 / 21.23 = 3.7 parts of tin

 In compound B
11.88 parts of oxygen combine with 88.12 parts of tin
1 part of oxygen combines with 88.13 / 11.88 = 7.4 parts of tin

 Thus, the mass of tin in compound A and B which combine with a fixed mass of oxygen are in the ratio of
3.7 : 7.4 or 1 : 2

 The data illustrates the law of multiple proportions
Law of Reciprocal proportions (or) law of equivalent proportions (Richter 1792 - 94)

“ If two elements combine separately with a third element, the mass ratio of the first two elements com-
bined with a fixed mass of the third element will be equal to or in simple ratio to the mass ratio of first two
elements in a compound formed by their direct combiantion”. For example :

i) Hydrogen combines with Sodium and chlorine to form two compounds. NaH and HCl respectively.

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Step II: Select the species undergoing oxidation and reduction and write them separately.
Step III: Balance the atoms of responsible element by simple counting.
Step IV: Balance the atoms of other elements by adding some molecule or ion in the proper side. The species

added should be according to the reaction and it should not creat a new change in O.S. In most of the
reactions, the other elements remained will be hydrogen or oxygen. They are balanced according to me-
dium of the reaction.

In acidic medium: Add one water molecule in the opposite side for each excess of oxygen atom. Add one H

ion in the opposite side for each excess of hydrogen atom.
In basic medium:

* Add one water molecule on the same side and two OH ions on opposite side for each excess of oxygen

atom.

* Add one OH ion on the same side and one water molecule on the opposite side for each excess of

hydrogen atom.

* The atoms may also be balanced by balancing them first in acidic medium and then replacing the H ions

suitably by OH ions.

Step V: Balance the charges in both process by adding proper number of electrons on the proper side.
Step VI: Make the total number of electrons lost and gained equal by multiplying with suitable numbers. Add

both the processes. It should be balanced reaction in ionic form.
Step VII: If the original reaction was in molecular form, convert it in molecular form.
CAP-48. Balance of the following reactions by ion electron method.

a)
4 2 4 2 4 4 2 2

KMnO H SO HCl K SO MnSO Cl H O     

Sol:
Step I: Ionic form of the given reaction is

2 2 2 2
4 4 4 4 2 2K MnO 2H SO H Cl 2K SO Mn SO Cl H O

                   

Step II: Oxidation:
0

2Cl C l



Reduction:
7

2
4Mn O Mn


 


Step III: Oxidation:
22Cl Cl

 

Reduction: 2
4MnO Mn
 

Step IV: Oxidation:
22Cl Cl

 

2
4 2MnO 8H Mn 4H O
    

Step V: Oxidation:
22Cl Cl 2e

  

Reduction: 2
4 2MnO 8H 5e Mn 4H O
     

Step VI: Oxidation:
2[2Cl Cl 2e] 5

   

Reduction: 2
4 2[MnO 8H 5e Mn 4H O] 2
      

______________________________________________________

2
4 2 22MnO 16H 10Cl 2Mn 8H O 5Cl
       

It is balanced reaction in ionic form.

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Step VII:
4 2 4 4 2 2 2 42KMnO 10HCl 3H SO 2MnSO 8H O 5Cl K SO      is the balanced reaction.

b)
2 3 2Cl OH Cl ClO H O

     

Step I: Oxidation:
2 3Cl ClO



Reduction:
2Cl Cl



Step II: Oxidation:
2 3Cl 2ClO



Reduction:
2Cl 2Cl



Step III: Oxidation:
2 3 2Cl 12OH 2ClO 6H O

   

Reduction:
2Cl 2Cl



Step IV: Oxidation:
2 3 2Cl 12OH 2ClO 6H O 10e

    

Reduction:
2Cl 2e 2Cl

 

Step V: Oxidation:
2 3 2Cl 12OH 2ClO 6H O 10e

    

Reduction:
2[Cl 2e 2Cl ] 5

  

________________________________________________

2 3 26Cl 12OH 2ClO 6H O 10Cl
     

or,
2 3 23Cl 6OH ClO 5Cl 3H O

      is the balanced reaction

Step III onwards may be replaced as

Step III: Oxidation:
2 2 3Cl 6H O 2ClO 12H

 
  

Reduction:
2Cl 2Cl



Step IV: Oxidation:
2 2 3Cl 6H O 2ClO 12H 10e

    

Reduction:
2Cl 2e 2Cl

 

Step V: Oxidation:
2 2 3[Cl 6H O 2ClO 12H 10e] 1

     

Reduction:
2[Cl 2e 2Cl ] 5

  

______________________________________________

2 2 36Cl 6H O 2ClO 10Cl 12H
     

or,
2 2 33Cl 3H O ClO 5Cl 6H

     

To remove H ion, add equal number of OH
 ions on both sides.

2 2 33Cl 3H O 6OH ClO 5Cl 6H 6OH
         

or,
2 2 3 23Cl 3H O 6OH ClO 5Cl 6H O

      

or,
2 3 23Cl 6OH ClO 5Cl 3H O

      is the balanced reaction.

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8. Estimation of Eq.mass of

9. Estimation Eq.mass of .

PART-9
(Significant Figures)

There is always some degree of uncertainty in every scientific measurement except in counting. The
uncertainty in measurement mainly depends upon two factors:
(i) Skill and accuracy of the observer,
(ii) Limitation of the measuring scale.
To indicate the precision of a measurement, scientists use the term significant figures. The significant
figures in a number are all the certain digits plus one doubtful digit. The number of significant figures gives
the information that except the digit at extreme right, all other digits are preciese or reproducible. For
example, mass of an object is 11.24g. This value indicates that actual mass of the object lies between
11.23g and 11.25g. Thus, one is sure of first three figures (1, 1 and 2) but the fourth figure is somewhat
inexact. The total significant figures in this number are four.

The following rules are observed in counting the number of significant figures in a given measured
quantity:
(i) All non-zero digits are significant. For example,

42.3 has three significant figures.
243.4 has four significant figures.
24.123 has five significant figures.

(ii) A zero becomes significant figure if it appears between two non-zero digits. For example.
5.03 has three significant figures.
5.604 has four significant figures.
4.004 has four significant figures.

(iii) Leading zeros or the zeros placed to the left of the number are never significant. For example,
0.543 has three significant figures.
0.045 has two significant figures.
0.006 has one significant figures.

(iv) Trailing zeros or the zeros placed to the right of the number are significant. For example,
433.0 has four significant figures.
433.00 has five significant figures.
343.000 has six significant figures.

(v) In exponential notation, the numerical portion gives the number of significant figures. For example.

has three significant figures.

has three significant figures.

(vi) The non-significant figures in the measurement are rounded off.
(a) If the figure following the last number to be retained is less than 5, all the unwanted figures are
discarded and the last number is left unchanged, e.g.,

5.6724 is 5.67 to three significant figures.
(b) If the figure following the last number to be retained is great than 5, the last figure to be retained is
increased by 1 unit and the unwanted figures are discarded, e.g.,

8.6526 is 8.653 to four significant figures.
(c) If the figure following the last number to be retained is 5, the last figure is increased by 1 only in case it
happens to be odd. In case of even number the last figure remains unchanged.

2.3524 is 2.4 to two significant figures.

7.4511 is 7.4 to two significant figures.

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Calculations Involving Significant Figures
In most of the experiments, the observations of various measurements are to be combined mathemati-

cally, i.e., added, subtracted, multiplied or divided as to achieve the final result. Since, all the observations in
measurements do not have the same precision, it is natural that the final result cannot be more precise than
the least precise measurement. The following two rules should be followed to obtain the proper number of
significant figures in any calculation.

Rule 1: The result of an addition or subtraction in the numbers having different precisions should be
reported to the same number of decimal places as are present in the number having the least number of
decimal places. The rule is illustrated by the following examples:
(a) 33.3 (has only one decimal place)

3.11
0.313

Sum 36.723 (answer should be reported to one decimal place)
Corret answer = 36.7
(b)3.1421

0.241
0.09 (has 2 decimal places)

Sum 3.4731 (answer should be reported to 2 decimal places)
Correct answer = 3.47
(c) 62.831 (has 3 decimal places)

-24.5492
Difference 38.2818 (answer should be reported to 3 decimla places after rounding off)
Correct answer = 38.282

Rule 2: The answer to a multiplication or division is rounded off to the same number of significatnt figures as is
possessed by the least precise term used in the calculation.
Examples are:
(a) 142.06

x0.23 (two significant figures)
32.6738 (answer should have two significant figures)

Correct answer = 33
(b)51.028

x1.31 (three significant figures)
66.84668

Correct answer = 66.8

(c)

Correct answer = 0.21
Note: (i) Same procedure is followed if an expression involves multiplication as well as division.

(ii) The presence of exact numbers in an expression does not affect the number of significant figures in the

answer.
Examples are:

(a) (b)

Correct answer = 0.0556 Correct answer = 44.8

_____

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